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2 years ago

Evaluate tan ( cos^-1 ( 1/8 ) ) , giving your answer as an exact value (no decimals)

1 answer:

5 0

$\bf cos^{-1}\left( \cfrac{1}{8} \right)=\theta \qquad \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{8}}\impliedby \textit{let's find the \underline{opposite}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{8^2-1^1}=b\implies \pm\sqrt{63}=b ~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\pm\sqrt{63}}}{\stackrel{adjacent}{1}} \\\\\\ ~\hspace{34em}$

$\bf \stackrel{\textit{keeping in mind that}}{tan\left(cos^{-1}\left( \frac{1}{8} \right) \right)}\implies tan(\theta )$

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Option A.) is correct $\triangle XYZ \sim \triangle WVZ$ by AA similarity. There are two angles that are congruent given by $\angle ZXY = \angle ZWV \hspace{0.2cm} and\hspace{0.2cm} \angle ZYX = \angle ZVW$ .

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Step-by-step explanation:

From the given figure we can say that

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3 years ago

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Nookie1986 [14]

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The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$